I want to prove the following:
$E$ is measurable, then $\forall \epsilon>0 \exists F(\text{closed}) \subset E \wedge m^*(E\sim F)< \epsilon$.
This is my attempt.
Let $\epsilon > 0$. Note that $E$ is measurable iff $E^c$ is measurable. Suppose that $m^*(E^c) < \infty$. Then, note that by the definition of the outer measure there exist an open cover of $E$, say $\{I_k\}_{k \in \mathbb{N}}$, thus:
$$E \subset \bigcup_{k \in \mathbb{N}}I_k \Rightarrow E^c \supset \bigcap_{k \in \mathbb{N}}I_k^c.$$
Define $F = \bigcap_{k \in \mathbb{N}}I_k^c$, note that $m^*(F) < \infty$. Now, by the definition of the outer measure, there exist a countable collection of open intervals, say $\{J_k\}_{k\in \mathbb{N}}$ that covers $F$ and
$$m^*(E^c) \leq \sum_{k\in \mathbb{N}}l(J_k) < m^*(F) + \epsilon \Rightarrow m^*(E^c\sim F) < \epsilon. \quad (1)$$
Now, assume that $m^*(E^c) = \infty$.Then there exist a countable collection of disjoint sets of finite outer measure such that $E^c = \bigsqcup_{k \in \mathbb{N}}E_k$. On the other hand, we have that since each of the $E_k$ has finite outer measure, $E_k \subset \mathcal{O}_k$ for $\mathcal{O}_k$ open and $$m^*(\mathcal{O}_k) \leq m^*(E_k) + \frac{\epsilon}{2^k}. \Leftrightarrow m^*(\mathcal{O}_k) - m^*(E_k) < \frac{\epsilon}{2^k}.$$ Now, we have that $E^c \subset \bigcup_{k\in \mathbb{N}}\mathcal{O}_k = \mathcal{O}$. Define $F = \mathcal{O}^c \Rightarrow \mathcal{O}^c \subset E^c$, we obtain: $$m^*(E^c\sim F) = m^*(E^c\sim \mathcal{O}^c) = m^*(\bigcup_{k\in \mathbb{N}}E_k \sim \bigcup_{k\in \mathbb{N}}\mathcal{O}_k) = m^*(\bigcup_{k\in \mathbb{N}}[E_k \sim \mathcal{O}_k]),$$ thus, $$m^*(E^c \sim F) \leq \sum_{k \in \mathbb{N}}m^*(E_k\sim \mathcal{O}_k) \leq \sum_{k \in \mathbb{N}}\frac{\epsilon}{k} = \epsilon.$$
Is the proof correct? I am not sure if the following argument is true:
Now, by the definition of the outer measure, there exist a countable collection of open intervals, say $\{J_k\}_{k\in \mathbb{N}}$ that covers $F$ and $$m^*(E^c) \leq \sum_{k\in \mathbb{N}}l(J_k) < m^*(F) + \epsilon \Rightarrow m^*(E^c\sim F) < \epsilon. \quad (1)$$